Question

Find $x$ for which ${\log }_{1/2}x>{\log }_{1/3}x$

• A. $\frac{1}{3}<x<\frac{1}{2}$
• B. $x>\frac{1}{2}$
• C. $0<x<1$
• D. $x<\frac{1}{3}$ and $x>\frac{1}{2}$

Answer 1

Answer: (C)

$0<x<1$

$\Rightarrow \frac{{\log }_{10}x}{{\log }_{10}(2{)}^{-1}}>\frac{{\log }_{10}x}{{\log }_{10}(3{)}^{-1}}$

$\Rightarrow \frac{-{\log }_{10}x}{{\log }_{10}2}>\frac{-{\log }_{10}x}{\log 3}$

$\Rightarrow \frac{{\log }_{10}x}{\log 3}-\frac{{\log }_{10}x}{{\log }_{10}2}>0$

$\Rightarrow {\log }_{10}x\left[\frac{\log 2-\log 3}{\log 2\log 3}\right]>0$

$\Rightarrow {\log }_{10}x<0(\because \log 2<\log 3)$

$\Rightarrow x<{10}^0\Rightarrow x<1$

and $x>0$ ($\because {\log }_{10}x$ is defined at $x>0$)

$\therefore 0<x<1$